The displacement of a particle in time t is given by $s = 2 t^{2} - 3 t + 1$ . The acceleration is

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Solution

The correct option is C 4
$s = 2 t^{2} - 3 t + 1; ∴ \frac{d s}{d t} = v = 4 t - 3$
At $t = 0 {(\frac{d s}{d t})}_{t .0} = - 3 = v_{1}$
Now, at t=1, we get ${(\frac{d s}{d t})}_{t = 1} = 4 - 3 = 1 = v_{2}$
Hence rate of velocity $= v_{2} - v_{1} = 1 - (- 3) = 4$
Aliter : Given that $s = 2 t^{2} - 3 t + 1$
$\frac{d s}{d t} = 4 t - 3$ (velocity), Again $\frac{d^{2} s}{d t^{2}} = 4$ (acceleration).

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